• Jan 9th 2007, 03:43 PM
gizmo97051
i need help figuring this out

Chip and Dale divided the peanuts that they collected in to 2 piles. ONe for doubles and one for triples. When finished shelling the peanuts you have 35 shells and 78 peanuts. HOw many shells were doubles.
• Jan 9th 2007, 06:00 PM
topsquark
Quote:

Originally Posted by gizmo97051
i need help figuring this out

Chip and Dale divided the peanuts that they collected in to 2 piles. ONe for doubles and one for triples. When finished shelling the peanuts you have 35 shells and 78 peanuts. HOw many shells were doubles.

Assume we have "d" doubles and "t" triples. Then we've got 35 peanuts (because that's how many shells we've got) so
$d + t = 35$

We get 2 peanuts for a double and 3 peanuts for a triple. Thus we've got
$2d + 3t = 78$

So solve the first equation for t:
$t = 35 - d$

And insert this value of t into the second equation:
$2d + 3(35 - d) = 78$

$2d + 105 - 3d = 78$

$-d = 78 - 105 = -27$

$d = 27$

So there were 27 in the doubles pile.

-Dan
• Jan 9th 2007, 06:10 PM
SysCoder
This is a systems of equations problem.

There are going to be to piles. One that has two nuts for each shell and the other three for each shell.

We will set the variable x to represent the shells that has tirpple and y for the shells that has double. So we can say every x is going to result in three peanuts (3x) and every y is going to result in 2 peanuts (2y). We can also reason and say 3x plus 2y equals the amount of peanuts.

First equation:
3x + 2y = 78

For the second equation we can say that x plus y equals the amount of shells

Second equation:
x + y = 35